Problem: Solve for $x$ : $ 3|x + 1| + 5 = -4|x + 1| + 6 $
Add $ {4|x + 1|} $ to both sides: $ \begin{eqnarray} 3|x + 1| + 5 &=& -4|x + 1| + 6 \\ \\ { + 4|x + 1|} && { + 4|x + 1|} \\ \\ 7|x + 1| + 5 &=& 6 \end{eqnarray} $ Subtract ${5}$ from both sides: $ \begin{eqnarray} 7|x + 1| + 5 &=& 6 \\ \\ { - 5} &=& { - 5} \\ \\ 7|x + 1| &=& 1 \end{eqnarray} $ Divide both sides by ${7}$ $ \dfrac{7|x + 1|} {{7}} = \dfrac{1} {{7}} $ Simplify: $ |x + 1| = \dfrac{1}{7}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 1 = -\dfrac{1}{7} $ or $ x + 1 = \dfrac{1}{7} $ Solve for the solution where $x + 1$ is negative: $ x + 1 = -\dfrac{1}{7} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& -\dfrac{1}{7} \\ \\ {- 1} && {- 1} \\ \\ x &=& -\dfrac{1}{7} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $7$ $ x = - \dfrac{1}{7} {- \dfrac{7}{7}} $ $ x = -\dfrac{8}{7} $ Then calculate the solution where $x + 1$ is positive: $ x + 1 = \dfrac{1}{7} $ Subtract ${1}$ from both sides: $ \begin{eqnarray} x + 1 &=& \dfrac{1}{7} \\ \\ {- 1} && {- 1} \\ \\ x &=& \dfrac{1}{7} - 1 \end{eqnarray} $ Change the ${ - 1}$ to an equivalent fraction with a denominator of $7$ $ x = \dfrac{1}{7} {- \dfrac{7}{7}} $ $ x = -\dfrac{6}{7} $ Thus, the correct answer is $x = -\dfrac{8}{7} $ or $x = -\dfrac{6}{7} $.